# Thread: Problem #9 - College Algebra

1. ## Problem #9 - College Algebra

Here we go for this week!
Can a four dimensional topological sphere have two or more inequivalent smooth structures?

Kidding! I'm kidding!

Here's the real one:

Solve the following system for (integers) x and y.
$\frac{1}{x} + \frac{1}{y} = -\frac{1}{2}$

$x^3 + y^3 = 7$

-Dan

2. ## Re: Problem #9 - College Algebra

Originally Posted by topsquark
Here we go for this week!
Can a four dimensional topological sphere have two or more inequivalent smooth structures?

Kidding! I'm kidding!

Here's the real one:

Solve the following system for (integers) x and y.
$\frac{1}{x} + \frac{1}{y} = -\frac{1}{2}$

$x^3 + y^3 = 7$

-Dan
The first question would send me back to school for years. In fact, I never took college algebra either: I so hated my calculus course at Columbia that I took no other math courses there.

$Solve\ for\ integer\ x\ and\ integer\ y\ given\ \dfrac{1}{x} + \dfrac{1}{y} = -\ \dfrac{1}{2}\ and\ x^3 + y^3 = 7.$

Spoiler:
$x^3 + y^3 = 7 \implies x = \sqrt[3]{7 - y^3}.$

$And\ \dfrac{1}{x} + \dfrac{1}{y} = -\ \dfrac{1}{2} \implies \dfrac{2}{x} + \dfrac{2}{y} = -\ 1 = \dfrac{- y }{y}\implies \dfrac{2}{x} = -\ \dfrac{y + 2}{y} \implies x = -\ \dfrac{2y}{y + 2}.$

$So\ -\ \dfrac{2y}{y + 2} = \sqrt[3]{7 - y^3} \implies \dfrac{8y^3}{y^3 + 6y^2 + 12y + 8} = y^3 - 7.$

Now that implies a polynomial of degree 6, but the coefficient of y6 is 1 so the integer root theorem applies. The constant term is - 56.

So if y is an integer $y = \pm 56, \pm 28, \pm 14, \pm 8, \pm 7, \pm 4, \pm 2, or\ \pm 1.$

Furthermore, we are looking for an integer x so y3 - 7 must be a perfect cube. Some numerical exploration shows that y = 2 works as does
y = - 1.

$y = 2 \implies x^3 + 2^3 = 7 \implies x^3 = -\ 1 \implies x = -\ 1.\ And\ \dfrac{1}{-1} + \dfrac{1}{2} = -\ \dfrac{1}{2}.$

$y = - 1 \implies x^3 + (-1)^3 = 7 \implies x^3 = 8 \implies x = 2.\ And\ \dfrac{1}{2} + \dfrac{1}{-1} = -\ \dfrac{1}{2}.$

Solutions are: $(-1, 2)\ or\ (2, -1).$

3. ## Re: Problem #9 - College Algebra

Another approach:
Spoiler:
Letting s = x + y and p = xy we get the two equations

$\frac{s}{p}=-\frac{1}{2}$

$s\left(s^2-3p\right)=7$

which in turn lead to

$s^3+6s^2=7$

$s=1$

$p=-2$

$x+y=1$

$xy=-2$

4. ## Re: Problem #9 - College Algebra

Thanks to all that helped out!

My solution is roughly the same as idea's and I have the unfortunate tendency to write out the first method I see, even if it isn't all that elegant. So here's my version in all it's multi-step glory!
Spoiler:

$\frac{1}{x} + \frac{1}{y} = -\frac{1}{2}$

Leads to
$\frac{x + y}{xy} = -\frac{1}{2}$

Leads to
$(x + y) = -\frac{1}{2}(xy)$

And, as is well known:
$(a + b)^3 = (a^3 + b^3) + 3(ab)(a + b)$

So we have:
$(x + y)^3 = (x^3 + y^3) + 3(xy)(x + y)$

So subbing in from this and the the initial condition $x^3 + y^3 = 7$
$\left ( -\frac{1}{2} (xy) \right ) ^3 = (7) + 3(xy) \left ( - \frac{1}{2} (xy) \right )$

With a little rearranging:
$(xy)^3 - 12(xy)^2 + 56 = 0$

Letting z = xy
$z^3 - 12z^2 + 56 = 0$

Since we are looking for integral solutions, let's try the rational root theorem. After sorting the list a bit to get all factors of 56 we find that z = -2 is the only rational solution. (The other two are irrational.) Now there is a detail we need to (quickly) look at. There is no a priori restriction here that says z must be an integer, only x and y. However if we look at the other two solutions for the z equation we can easily see that z = -2 is the only way to go.

So $z = xy = -2$ gives $y = -\frac{2}{x}$

Thus
$\frac{1}{x} + \frac{1}{y} = \frac{1}{x} + \frac{1}{-\frac{2}{x}} = -\frac{1}{2}$

Which leads to
$x^2 + x - 2 = 0$

This has two solutions: (x,y) = (-2. 1), (1, -2).

-Dan